2 Answers
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If there does not exist such $E,F$, then $\lambda$ would be finitely additive. We now follow the proof from the book by using the equivalence set $S$ consisting of one element each in $[0,1]/\mathbb{Q}$. Let $q_{n}$ be an eumeration of rationals in $(-1,1)$ and let $E_{n}=S+q_{n}$. We now conclude that
$$1=\lambda^{*}(0,1)\le \lambda^{*}(\bigcup E_{n})\le \sum \lambda^{*}(E_{n})=\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\lambda^{*}(E_{k})=\lim_{n\rightarrow \infty}\lambda^{*}(\bigcup^{n}_{k=1}E_{k})\le \lambda^{*}(-1,2)=3$$

But on the other hand $$\sum^{n}_{k=1}\lambda^{*}E_{k}=\sum^{n}_{k=1}\lambda^{*}(S+q_{k})=\sum^{n}_{k=1}\lambda^{*}(S+q_{k})=\sum^{n}_{k=1}\lambda^{*}(S)=n\lambda^{*}(S)$$ This implies in particular $1\le n\lambda^{*}S\le 3$, which is impossible for any finite number $\lambda^{*}(S)$. We thus concluded the proof. And such $E,F$ must exist by contradiction.

A slightly more explicit construction is a Bernstein set: a subset $B \subseteq \mathbb{R}$ such that given any uncountable closed $F \subseteq \mathbb{R}$ both $F \cap B$ and $F \setminus B$ are non-empty. (The construction of such a set follows from (a fragment of) the Axiom of Choice.)

Suppose that $B$ is a Bernstein subset of $[0,1]$ (meaning $B \subseteq [0,1]$ and given any uncountable closed $F \subseteq [0,1]$ both $F \cap B$ and $F \setminus B$ are non-empty). Note that if $B \subseteq \bigcup_{i =1}^\infty I_i$ where each $I_i$ is an open interval, then $F = [0,1] \setminus \bigcup_{i=1}^\infty I_i$ is a closed subset of $[0,1]$ and $F \cap B = \emptyset$. Then it must be that $F$ is countable and as $[0,1] \subseteq F \cup \bigcup_{i=1}^\infty I_i$ it follows that $$\textstyle{\sum_{i=1}^\infty} \mathrm{length} ( I_i ) = \lambda^* \left( \textstyle{\bigcup_{i=1}^\infty} I_i \right) \geq 1.$$ We may then conclude that $\lambda^* ( B ) = 1$. The same argument will show that $\lambda^* ( [0,1] \setminus B ) = 1$.